Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:

(y^3)y''=1

So far, using a substitution of p=y', I have managed to do:

(y^3)p(dp/dy)=1

p dp = 1/(y^3) dy

(p^2)/2 = -1/(2y^2) + c

p^2 = -1/(y^2) + c

Can someone please confirm if my working so far is correct or incorrect and where to go from here.

Thanks